Question: The area of a circle is increasing at a rate of $8\pi$ square meters per hour. At a certain instant, the area is $36\pi$ square meters. What is the rate of change of the circumference of the circle at that instant (in meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $4\sqrt{2}$ (Choice B) B $\dfrac{4\pi}{3}$ (Choice C) C $12\pi$ (Choice D) D $\dfrac{\pi}{6}$
Solution: Setting up the math Let... $r(t)$ denote the circle's radius at time $t$, $A(t)$ denote the circle's area at time $t$, and $C(t)$ denote the circle's circumference at time $t$. We are given that $A'(t)=8\pi$, We are also given that $A(t_0)=36\pi$ for a specific time $t_0$. We want to find $C'(t_0)$. Relating the measures $C(t)$ and $r(t)$ relate to each other through the formula for the circumference of a circle: $C(t)=2\pi r(t)$ We can differentiate both sides to find an expression for $C'(t)$ : $C'(t)=2\pi r'(t)$ $A(t)$ and $r(t)$ relate to each other through the formula for the area of a circle: $A(t)=\pi[r(t)]^2$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=2\pi r(t)r'(t)$ Using the information to solve Let's plug ${A(t_0)}={36\pi}$ into the expression for $A(t_0)$ : $\begin{aligned} {A(t_0)}&=\pi[r(t_0)]^2 \\\\ {36\pi}&=\pi[r(t_0)]^2 \\\\ 36&=[r(t_0)]^2 \\\\ {6}&={r(t_0)} \end{aligned}$ Let's plug ${A'(t_0)}={8\pi}$ and ${r(t_0)}={6}$ into the expression for $A'(t_0)$ : $\begin{aligned} {A'(t_0)}&=2\pi{r(t_0)}r'(t_0) \\\\ {8\pi}&=2\pi({6})r'(t_0) \\\\ C{\dfrac{2}{3}}&=C{r'(t_0)} \end{aligned}$ Now let's plug $C{r'(t_0)}=C{\dfrac{2}{3}}$ into the expression for $C'(t_0)$ : $\begin{aligned} C'(t_0)&=2\piC{r'(t_0)} \\\\ &=2\pi\left(C{\dfrac{2}{3}}\right) \\\\ &=\dfrac{4\pi}{3} \end{aligned}$ In conclusion, the rate of change of the circumference of the circle at that instant is $\dfrac{4\pi}{3}$ meters per hour. Since the rate of change is positive, we know that the circumference is increasing.